Two point charges + 9 electrons and protons electrons are kept at a distance is from each other where should we place a third charge Q on the line joining the two charges so that it may be in equilibrium
Answers
Answer:
The charge on the first particle is +9e
The charge on the second particle is +e
The distance between both the particle is L=0.16 m
Let the distance x from the first particle to which the third particle of charge q will be in equilibrium
The electric force between first and third charge is
F_{1}=\dfrac{9keq}{x^2}F1=x29keq
The electric force between the second and third particle is
\begin{lgathered}F_{2}=\dfrac{keq}{(L-x)^2}\\\end{lgathered}F2=(L−x)2keq
As the net force on the third charge is zero
So equating both the forces
\begin{lgathered}F_{1}=F_{2}\\\dfrac{9keq}{x^2}=\dfrac{keq}{(L-x)^2}\\\dfrac{9}{x^2}=\dfrac{1}{(L-x)^2}\\(L-x)^2=9x^2\end{lgathered}F1=F2x29keq=(L−x)2keqx29=(L−x)21(L−x)2=9x2
Taking square root at both sides we get,
\begin{lgathered}\sqrt{(L-x)^2}=\sqrt{9x^2}\\L-x=3x\\4x=L\\4x=0.16\\x=\dfrac{0.16}{4}\\x=0.04\ m\end{lgathered}(L−x)2=9x2L−x=3x4x=L4x=0.16x=40.16x=0.04 m